Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $n = \dfrac{2r - 20}{r + 4} \times \dfrac{r^2 + 5r + 4}{r^2 - 10r} $
Explanation: First factor the quadratic. $n = \dfrac{2r - 20}{r + 4} \times \dfrac{(r + 4)(r + 1)}{r^2 - 10r} $ Then factor out any other terms. $n = \dfrac{2(r - 10)}{r + 4} \times \dfrac{(r + 4)(r + 1)}{r(r - 10)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ 2(r - 10) \times (r + 4)(r + 1) } { (r + 4) \times r(r - 10) } $ $n = \dfrac{ 2(r - 10)(r + 4)(r + 1)}{ r(r + 4)(r - 10)} $ Notice that $(r - 10)$ and $(r + 4)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ 2\cancel{(r - 10)}(r + 4)(r + 1)}{ r\cancel{(r + 4)}(r - 10)} $ We are dividing by $r + 4$ , so $r + 4 \neq 0$ Therefore, $r \neq -4$ $n = \dfrac{ 2\cancel{(r - 10)}\cancel{(r + 4)}(r + 1)}{ r\cancel{(r + 4)}\cancel{(r - 10)}} $ We are dividing by $r - 10$ , so $r - 10 \neq 0$ Therefore, $r \neq 10$ $n = \dfrac{2(r + 1)}{r} ; \space r \neq -4 ; \space r \neq 10 $